这对我来说是非常矛盾的，让我尽力解释。这就像要计算的2层。第一层是我需要得到总A，总B，总C和总D。但是，要获得总A，我需要计算一些列，以及总B，C和D.第二层是显示并获得总A，B，C，D的总和。这是一堆表。
第一张表。
tbl_criteria
------------------------
crit_id | criteria_name
16 | sports
17 | formal
18 | talent
19 | nothing
tbl_criteria有子标准
tbl_sub_criteria
----------------------
sub_crit_id | crit_id | sub_crit_name
22 | 16 | originality
23 | 16 | audience Impact
24 | 18 | Appeal
25 | 18 | Stage Presence
第三表评委。
tbl_judges
------------------------
judge_id | judge_name
61 | first
62 | second
63 | third
参赛者表让2名选手
tbl_cotestant
-----------------------------------------------
con_id | contestant_number | contestant_name |
1 | 1 | john |
2 | 2 | sy |
最后一张表，这是构造的表
tbl_score
--------------------------------------------------
score_id | crit_id | sub_crit_id | judge_id | con_id | contestant_number | score
1 | 16 | 22 | 61 | 1 | 1 | 25
2 | 16 | 22 | 61 | 2 | 2 | 25
3 | 16 | 22 | 62 | 1 | 1 | 25
4 | 16 | 22 | 62 | 2 | 2 | 73
5 | 16 | 22 | 63 | 1 | 1 | 70
6 | 16 | 22 | 63 | 2 | 2 | 80
7 | 16 | 23 | 61 | 1 | 1 | 25
8 | 16 | 23 | 61 | 2 | 2 | 25
9 | 16 | 23 | 62 | 1 | 1 | 25
10 | 16 | 23 | 62 | 2 | 2 | 73
11 | 18 | 23 | 63 | 1 | 1 | 70
12 | 16 | 23 | 63 | 2 | 2 | 80
13 | 18 | 24 | 61 | 1 | 1 | 25
14 | 18 | 24 | 61 | 2 | 2 | 25
15 | 18 | 24 | 62 | 1 | 1 | 25
16 | 18 | 24 | 62 | 2 | 2 | 73
17 | 18 | 24 | 63 | 1 | 1 | 70
18 | 18 | 24 | 63 | 2 | 2 | 80
19 | 18 | 25 | 61 | 1 | 1 | 25
20 | 18 | 25 | 61 | 2 | 2 | 25
21 | 18 | 25 | 62 | 1 | 1 | 25
22 | 18 | 25 | 62 | 2 | 2 | 73
23 | 18 | 25 | 63 | 1 | 1 | 70
24 | 18 | 25 | 63 | 2 | 2 | 80
25 | 17 | null | 61 | 1 | 1 | 25
26 | 17 | null | 61 | 2 | 2 | 25
27 | 17 | null | 62 | 1 | 1 | 25
28 | 17 | null | 62 | 2 | 2 | 73
29 | 17 | null | 63 | 1 | 1 | 70
30 | 17 | null | 63 | 2 | 2 | 80
31 | 19 | null | 61 | 1 | 1 | 25
32 | 19 | null | 61 | 2 | 2 | 25
33 | 19 | null | 62 | 1 | 1 | 25
34 | 19 | null | 62 | 2 | 2 | 73
35 | 19 | null | 63 | 1 | 1 | 70
36 | 19 | null | 63 | 2 | 2 | 80
第一层输出是这样的，得到总A，B，C，D。
总计应显示为
(criteria 16 has two sub-criterias 22, 23 , that means it will be x2)
con_num | contestant_name | 16_judge_61 | 16_judge_62 | 16_judge_63 | total a
1 | john | 50 | 25 | 140 | 215
2 | sy | 50 | 146 | 160 | 365
上面的表格中，John在Criteria number 16（crit_id 16）中共得到215个。同样，sy在标准编号16中总共有365个。
所以，我的表16,17,18,19中有4个标准。这是我的问题，这意味着我需要逐个查询以获得每个总输出，这就像
con_num | contestant_name | 17_judge_61 | 17_judge_62 | 17_judge_63 | total b
1 | john | 25 | 25 | 70 | 120
2 | sy | 25 | 73 | 80 | 178
(criteria 18 has a sub criteria 24,25 that means it will x2)
con_num | contestant_name | 18_judge_61 | 18_judge_62 | 18_judge_63 | total c
1 | john | 50 | 50 | 140 | 240
2 | sy | 50 | 146 | 160 | 356
con_num | contestant_name | 19_judge_61 | 19_judge_62 | 19_judge_63 | total d
1 | john | 25 | 25 | 70 | 120
2 | sy | 25 | 73 | 80 | 178
我需要在一个查询中执行此操作，我只能逐个获得总计A，B C.但我需要一个执行此类输出的查询。我怎样才能实现这一输出？
total_a，b，c，d相当于crit_id 16,17,18,19
con_num | contestant_name | 16total_a | 17total_b | 18total_c | 19total_d | Grand_total
1 | john | 215 | 120 | 240 | 120 | 695
2 | jy | 365 | 178 | 356 | 178 | 1077
这个查询，我还在学习逻辑
SELECT DISTINCT(a.contestant_number) as con_num, a.contestant_name,
//Getting first the sum of total a, Notice criteria 16 has a sub criteria 22, 23
SUM(CASE WHEN s.crit_id='16' AND s.sub_crit_id = 22 AND s.judge_id='61' THEN s.score END) as 16_judge_61,
SUM(CASE WHEN s.crit_id='16' AND s.sub_crit_id = 22 AND s.judge_id='62' THEN s.score END) as 16_judge_62,
SUM(CASE WHEN s.crit_id='16' AND s.sub_crit_id = 22 AND s.judge_id='63' THEN s.score END) as 16_judge_63,
SUM(CASE WHEN s.crit_id='16' AND s.sub_crit_id = 23 AND s.judge_id='61' THEN s.score END) as 16_judge_61,
SUM(CASE WHEN s.crit_id='16' AND s.sub_crit_id = 23 AND s.judge_id='62' THEN s.score END) as 16_judge_62,
SUM(CASE WHEN s.crit_id='16' AND s.sub_crit_id = 23 AND s.judge_id='63' THEN s.score END) as 16_judge_63,
SUM(CASE WHEN s.crit_id='16' AND s.judge_id in (61, 62, 63) THEN s.score END) as 'total a'
//Criteria 17 has no sub criteria
SUM(CASE WHEN s.crit_id='17' AND s.judge_id='61' THEN s.score END) as 16_judge_61,
SUM(CASE WHEN s.crit_id='17' AND s.judge_id='62' THEN s.score END) as 16_judge_62,
SUM(CASE WHEN s.crit_id='17' AND s.judge_id='63' THEN s.score END) as 16_judge_63,
SUM(CASE WHEN s.crit_id='17' AND s.judge_id in (61, 62, 63) THEN s.score END) as 'total b'
//And iterate again for the 2 more criteria
//
FROM tbl_score s
INNER JOIN tbl_contestant a ON s.contestant_number = a.contestant_number
INNER JOIN tbl_judges j ON j.judge_id = s.judge_id
WHERE c.gender = 'male' and c.con_id = s.con_id
GROUP BY s.contestant_number
ORDER By `Grand toal' DESC";
http://sqlfiddle.com/#!9/cfe90/1
获得总数的总和 http://sqlfiddle.com/#!9/9efa5/1
答案 0 :(得分：1)
将您的查询放在子查询中，然后将列添加到一起以获得总计。
SELECT x.*, `total a` + `total b` + `total c` AS `Grand Total`
FROM (put your query here) AS x